Answer :
The roast will be at 150°F approximately 84.39 minutes after being put into the oven.
To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. The differential equation representing this law is:
[tex]\[ \frac{{dT}}{{dt}} = k(T - T_a) \][/tex]
Where:
- [tex]\( \frac{{dT}}{{dt}} \)[/tex] is the rate of change of temperature with respect to time.
- ( T ) is the temperature of the object.
- [tex]\( T_a \)[/tex] is the ambient temperature (oven temperature in this case).
- ( k ) is a constant of proportionality.
To solve this differential equation, we need to find ( k ). We're given that the roast is initially at 50°F and placed into a 375°F oven, so[tex]\( T(0) = 50 \)[/tex] and[tex]\( T_a = 375 \).[/tex]
[tex]\[ \frac{{dT}}{{dt}} = k(50 - 375) = -325k \][/tex]
Given that the temperature is 125°F after 75 minutes, we can plug in these values to find ( k ):
[tex]\[ 125 = 50 + (375 - 50)e^{-325k \times 75} \][/tex]
Now, let's solve for ( k ):
[tex]\[ 75 = 325k \times 75 \][/tex]
[tex]\[ k = \frac{1}{325} \][/tex]
Now, we have the value of ( k ), so we can use it to solve for the time it takes for the roast to reach 150°F.
[tex]\[ 150 = 50 + (375 - 50)e^{-\frac{1}{325} \times t} \][/tex]
[tex]\[ 100 = 325e^{-\frac{1}{325} \times t} \][/tex]
[tex]\[ e^{-\frac{1}{325} \times t} = \frac{100}{325} \][/tex]
[tex]\[ -\frac{1}{325} \times t = \ln\left(\frac{100}{325}\right) \][/tex]
[tex]\[ t = -325 \times \ln\left(\frac{100}{325}\right) \][/tex]
[tex]\[ t \approx 84.39 \text{ minutes} \][/tex]
So, the roast will be at 150°F approximately 84.39 minutes after being put into the oven.
